1


I've read in some places that the following proof from Cours d'Analyse of Cauchy is not correct, but I can't find the mistake.

Theorem: if the difference $f(x+1)-f(x)$ converges towards a certain limit $k$, for increasing values of $x$, then the fraction $$\frac{f(x)}{x}$$ converges at the same time towards the same limit.

Proof:
First suppose that the quantity $k$ has a finite value, and denote by ε a number as small as we wish. Because the increasing values of $x$ make the difference $$f(x+1)-f(x)$$converge towards the limit $k$, we can give the number $h$ a value large enough that, when $x$ is equal to or greater than $h$, the difference in question is always contained between the limits $k-ε$ and $k+ε$.

Given this, if we denote by n any integer number, $$f(h+1)-f(h)$$ $$f(h+2)-f(h+1)$$

$$f(h+n)-f(h+n-1)$$
Consquently their arithmetic mean $$\frac{f(h+n)-f(h)}{n}$$ is contained between $k-ε$ and $k+ε$. Thus we have: $$\frac{f(h+n)-f(h)}{n}=k+α$$where α is a quantity contained between the limits −ε and +ε. Now let $h+n=x$. Therefore $$\frac{f(x)-f(h)}{x-h}=k+α$$ and thus we conclude $$f(x) = f (h) + (x−h) (k+α)$$$$\frac{f(x)}{x}=\frac{f(h)}{x}+(1-\frac{h}{x})(k+α)$$ Moreover, to make the value of $x$ increase indefinitely, it suffices to make the integer number $n$ increase indefinitely without changing the value of $h$. Consequently, let us suppose that in equation we consider $h$ as a constant quantity and $x$ as a variable quantity which converges towards the limit ∞.

The quantities $\frac{f(h)}{x}$ and $\frac{h}{x}$ contained in the right-hand side, converge towards the limit zero, and the righthand side itself converges towards a limit of the form $k+α$ where α is always contained between −ε and +ε.

Thus the ratio $\frac{f(x)}{x}$ has for its limit a quantity contained between $k − ε$ and $k + ε$. This conclusion remains true however small the number ε may be, and as a result the limit in question is precisely the quantity $k$.
In other words: $lim \frac{f(x)}{x} = k = lim[ f (x+1)− f (x)]$.
Q.E.D.

Neo B.
  • 23
  • 3
  • Where did you read that this wasn't correct, exactly? – Matthew Towers Aug 16 '18 at 18:09
  • 2
    If you fix $h$, then not every large enough $x$ can be represented as $h+n$, where $n$ is a positive integer. – Litho Aug 16 '18 at 18:15
  • 1
    Your "every $x$" is actually a very particular sequence $x_n$ with $x_{n+1} = x_n + 1$. – user251257 Aug 16 '18 at 18:25
  • @m_t_ I read it in a text about calculus' history. The important thing is that there's a counterexample of the "theorem", the function $\frac{1}{1-x+[x]}$. – Neo B. Aug 16 '18 at 18:35
  • 2
    The theorem is wrong. The correct version requires that $f$ is bounded on every bounded interval. See https://math.stackexchange.com/q/192963/72031 – Paramanand Singh Aug 17 '18 at 01:12

1 Answers1

2

This is NOT a valid proof. For $y\in (0,1]$ and $n\in \Bbb N$ let $f_n(y)=\frac {f(y+n)}{y+n}.$ The proof only shows that $\lim_{n\to \infty}f_n(y)=k$ for each $y\in (0,1].$

But the assertion that $\lim_{x\to \infty}\frac {f(x)}{x}=k$ is equivalent to the statement that $f_n(y)\to k$ uniformly on $(0,1].$

The counter-example $f(x)=\frac {1}{1-x+[x]}$ you cite in a comment is good. We have $f(x+1)-f(x)=0$ for all $x.$ We have $\lim_{n\to \infty} f_n(y)=0$ for each $y\in (0,1].$

But $f_n(1-\frac {1}{n})=$ $\frac {n}{n+1-\frac {1}{n}}>$ $\frac {n}{n+1}\geq \frac {1}{2}$ so the convergence of $f_n$ to $0$ on $(0,1]$ is not uniform.

DanielWainfleet
  • 59,529
  • 1
    If you read the original works of Cauchy (and maybe some of his contemporaries) there are some things that don't seem to make sense, because "convergent" might implicitly mean "absolute convergent" or "uniform convergent", as the terminology had not been standardized. But this "proof" is not a terminological mis-understanding. It's just wrong. – DanielWainfleet Aug 17 '18 at 19:04